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Power Output of a Turbine (Rotational Energy)

  1. Feb 6, 2017 #1
    1. The problem statement, all variables and given/known data
    A Verdant Power water turbine (a “windmill” in water) turns in the East River near New York City. Its propeller is 2.5 m in radius and spins at 32 rpm when in water that is moving at 2.0 m/s. The rotational inertia of the propeller is approximately 3.0 kg∙m^2. Determine the kinetic energy of the turbine and the electric energy in joules that it could provide in one day if it is 100% efficient at converting its kinetic energy into electric energy.

    2. Relevant equations
    tau=Fr
    tau=I(alpha)
    KE=0.5I(omega)^2
    P=tau(omega)
    W=tau(theta)

    3. The attempt at a solution
    I can find the rotational kinetic energy using the third equation in the list, and I get 17 J. But I'm having trouble finding the power produced by the wheel, and I wanted to do that because then I can do power x time to find the energy produced in a day. I can't find the torque on the wheel because the wheel is moving at constant velocity and I have no information about friction. I feel like I should use the rotational KE in some way, but dividing the rotational KE by one day to get power won't make sense because they I'd just have to multiply it by one day and be back where I started.

    Any help appreciated. :)
     
  2. jcsd
  3. Feb 6, 2017 #2

    jbriggs444

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    At first glance, it sounds like a trick question. "How much electrical energy can a turbine deliver if it converts its rotational kinetic energy to electrical energy?"

    Answer: "All of it".

    Another reading is that the intended question asks about the energy that the turbine can extract from the energy in the flowing water. The naive answer would be the same, "All of it". Unfortunately, that answer is not correct. There is a problem. If you could extract all of the energy from the flowing water then the water coming out of the turbine would not be moving. There could then be no flow at all. This difficulty imposes a limit on the maximum achievable efficiency of a turbine. https://en.wikipedia.org/wiki/Betz's_law.
     
    Last edited: Feb 7, 2017
  4. Feb 6, 2017 #3
    I understand how you're approaching this question from a conceptual standpoint, but the way the question is worded makes it sound like it's looking for a numerical answer, and I'm not sure how to arrive there.

    I had another thought -- perhaps this question requires some knowledge of how turbines work. If the radius of the turbine is 2.5 m, then is the water falling a distance of 5.0 m? If so, I can calculate the change in gravitational potential energy of the water, except that I don't have any way to calculate how quickly the water is traveling through the turbine in kg/s, which I think I would need to know.
     
  5. Feb 6, 2017 #4

    haruspex

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    Just to elaborate on jbriggs' answer...
    You are told the turbine rotates at a constant speed, so none of its KE is being converted to electrical energy, and its efficiency at such a conversion is irrelevant. We must therefore interpret that as meaning all the work done on it is converted efficiently. But, as you say, we do not know the torque, so we do not know how much work is being done on it.
    Certainly the information about its moment of inertia and rate of rotation are irrelevant to the power developed. Given the water velocity, the other useful piece of information would be the pressure drop.
     
  6. Feb 6, 2017 #5

    CWatters

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    Its a slightly badly worded question but you can calculate the power available at least.
     
  7. Feb 6, 2017 #6

    haruspex

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    At the least, it is very badly worded, and I'm not so sure the available power can be calculated. The flow velocity is given as a constant. Indeed, in a pipe of constant radius and an incompressible fluid, the linear velocity is necessarily constant.
    I'm certainly no expert on this topic, so I could have this wrong, but from looking at Bernoulli's equation it seems to me that the power comes from the pressure difference multiplied by the area and the velocity, and we have no way to know the pressure difference.

    Edit: following Nidum's link, I see this is a run-of-river installation. Even so, there needs to be a pressure differential. See e.g. http://www.climatetechwiki.org/technology/ror.

    Betz' Law is for compressible flows, I believe.
    You know the linear speed and the cross-sectional area.
     
    Last edited: Feb 6, 2017
  8. Feb 6, 2017 #7

    Nidum

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    This problem relates to a submerged axial flow turbine .

    Not very accurate but mass flow of water could be estimated from the given flow velocity and the turbine diameter .
     
  9. Feb 7, 2017 #8
    Extensive googling has revealed that the instructor accidentally left off a sentence at the end of the question, therefore there wasn't enough information in the problem to solve it. Case closed.
     
  10. Feb 7, 2017 #9

    haruspex

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    Yes, it seems to be a popular question. Here's the missing info. It is quite bizarre and misleading:
    "Assume that the energy delivered per revolution is equal to the rotational kinetic energy of the turbine. "
     
  11. Feb 7, 2017 #10

    jbriggs444

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    Wow. Bizarre and misleading indeed. Thank you for tracking that down and adding it to the thread.
     
  12. Feb 7, 2017 #11

    jbriggs444

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    I finally got around to looking this one up. Betz' law holds for incompressible flows. Flowing water is suitably incompressible, of course. Blowing wind is also approximately incompressible in most relevant applications.
     
  13. Feb 7, 2017 #12

    haruspex

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    Ok, thanks.
     
  14. Feb 8, 2017 #13

    haruspex

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    I followed up on the Betz law derivation. It assumes that the energy comes from the loss of KE in the stream. That would apply if you simply lowered a turbine into a wider stream of flowing water, so that there is essentially no pressure difference each side. But even run-of-river turbines do not work that way. They use a leet from further upstream so as to get a pressure head. The turbine itself is in a tubular enclosure, so there can be no velocity drop through it.
     
  15. Feb 8, 2017 #14

    jbriggs444

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